Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{3z}{z^2 + 9z} \div \dfrac{-10}{6(z + 9)} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{3z}{z^2 + 9z} \times \dfrac{6(z + 9)}{-10} $ When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ 3z \times 6(z + 9) } { (z^2 + 9z) \times -10 } $ $ a = \dfrac {3z \times 6(z + 9)} {-10 \times z(z + 9)} $ $ a = \dfrac{18z(z + 9)}{-10z(z + 9)} $ We can cancel the $z + 9$ so long as $z + 9 \neq 0$ Therefore $z \neq -9$ $a = \dfrac{18z \cancel{(z + 9})}{-10z \cancel{(z + 9)}} = -\dfrac{18z}{10z} = -\dfrac{9}{5} $